Answer:
<u>1)Thus based on the diagram , the values of principle quantum number that can be assigned to the electrons are : 2,3 and 4.</u>
<u>2)This quantum number tells us the shell in which an electron is present and how far it is roughly from the nucleus.</u>
Explanation:
According to the rules of electron filling in subshells in quantum mechanics , <em>Aufbau's principle states that :</em>
<em>"An electron must be filled in a lower energy state or subshell and then higher. If they have the same energy , the one with lesser principal quantum number will be filled first."</em>
Thus ,
The series of filling goes like:
![1.1s\\2.2s 2p\\3.3s3p4s\\4.3d4p5s\\...](https://tex.z-dn.net/?f=1.1s%5C%5C2.2s%202p%5C%5C3.3s3p4s%5C%5C4.3d4p5s%5C%5C...)
Thus , the subshells in the given information have 1,3 and 1 orbitals respectively. Thus it has to be
which is possible in two cases :
![1.2s2p3s\\2.3s3p4s](https://tex.z-dn.net/?f=1.2s2p3s%5C%5C2.3s3p4s)
1)Thus based on the diagram , the values of principle quantum number that can be assigned to the electrons are : 2,3 and 4.
2)This quantum number tells us the shell in which an electron is present and how far it is roughly from the nucleus.
The sensitivity of electronic balance is 0.3 mg
Converting 0.3 mg to g:
g
So, the mass of a substance weighed on the balance can be measured accurately up to four decimal places.
Given the mass of sample is 1.2300 g. The number of significant figures in the measurement will be 5 because the balance can measure the mass of a sample accurately up to four decimal points, all the digits in the measurement will be significant.
<u>Answer:</u> The concentration of
in the solution is ![1.87\times 10^{-14}M](https://tex.z-dn.net/?f=1.87%5Ctimes%2010%5E%7B-14%7DM)
<u>Explanation:</u>
The given cell is:
![Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)](https://tex.z-dn.net/?f=Pt%28s%29%7CH_2%28g.1atm%29%7CH%5E%2B%28aq.%2C1.0M%29%7C%7CAu%5E%7B3%2B%7D%28aq%2C%3FM%29%7CAu%28s%29)
Half reactions for the given cell follows:
<u>Oxidation half reaction:</u>
( × 3)
<u>Reduction half reaction:</u>
( × 2)
<u>Net reaction:</u> ![3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)](https://tex.z-dn.net/?f=3H_2%28s%29%2B2Au%5E%7B3%2B%7D%28%3FM%29%5Crightarrow%206H%5E%7B%2B%7D%281.0M%29%2B2Au%28s%29)
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the
of the reaction, we use the equation:
![E^o_{cell}=E^o_{cathode}-E^o_{anode}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7Bcathode%7D-E%5Eo_%7Banode%7D)
Putting values in above equation, we get:
![E^o_{cell}=1.50-0=1.50V](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3D1.50-0%3D1.50V)
To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:
![E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E6%7D%7B%5BAu%5E%7B3%2B%7D%5D%5E2%7D)
where,
= electrode potential of the cell = 1.23 V
= standard electrode potential of the cell = +1.50 V
n = number of electrons exchanged = 6
![[Au^{3+}]=?M](https://tex.z-dn.net/?f=%5BAu%5E%7B3%2B%7D%5D%3D%3FM)
![[H^{+}]=1.0M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D1.0M)
Putting values in above equation, we get:
![1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})](https://tex.z-dn.net/?f=1.23%3D1.50-%5Cfrac%7B0.059%7D%7B6%7D%5Ctimes%20%5Clog%28%5Cfrac%7B%281.0%29%5E6%7D%7B%5BAu%5E%7B3%2B%7D%5D%5E2%7D%29)
![[Au^{3+}]=1.87\times 10^{-14}M](https://tex.z-dn.net/?f=%5BAu%5E%7B3%2B%7D%5D%3D1.87%5Ctimes%2010%5E%7B-14%7DM)
Hence, the concentration of
in the solution is ![1.87\times 10^{-14}M](https://tex.z-dn.net/?f=1.87%5Ctimes%2010%5E%7B-14%7DM)
A molecule is a matter that moves around fast
Answer:
1. Convert the mass of A to moles dividing in molar mass
2. Convert the moles of A to moles of D (2mol A = 3mol D)
3. Convert the moles of D to mass multiplying by its molar mass
Explanation:
As you can see in the equation, 2moles of A produce 3 moles of D, the equation is:
2mol A = 3mol D
But in the problem you are having the mass of A and you need the mass of D. That means you require to convert the mass of A to moles. With the moles of A you can find the moles of D and, as last, you need to convert the moles of A to mass.
To convert mass to moles or vice versa you must use <em>Molar Mass. </em>That means, the answer is:
<h3>
1. Convert the mass of A to moles dividing in molar mass</h3><h3>
2. Convert the moles of A to moles of D (2mol A = 3mol D
)</h3><h3>
3. Convert the moles of D to mass multiplying by its molar mass</h3><h3 />